The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of `sqrt(3gl)`. Find the angle rotated by the string before it becomes slack.

Answer

A, C

Explanation

`V=sqrt(3gl)`
`1/2mv^2-1/2mu^2=-mgh`
`v^2=u^2-2g(l+lcostheta)`……….i
`Again `(mv^2)/l=mgcostheta`
`v^2=lg costheta`

From equation i and ii we get
`3gl-2gl-2gl cos theta=glcostheta`
`3costheta=1`
`=cos^-1(-1/3)`